statics problem (:

- anonymous

statics problem (:

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- anonymous

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- anonymous

im confused as to what forces are acting on points a and c.

- anonymous

Take components

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## More answers

- anonymous

okay, so i did the sum of F in the y direction.. = Ay + Cy - W - T = 0
and then the sum of F in the x direction... = Ax + Cx - Fb = 0
then i did the moment about A and got Fb = 8.9, but i am stuck there.

- anonymous

oh and Fb is acting in the negative x direction on point B.

- anonymous

At point A, it's the Normal reaction from Ground and the Tension from the string... at C it's just the string, is Pulley Mas-less and friction-less?

- anonymous

i assumeee the pulley has no mass nor friction.

- anonymous

wait so how many unknowns do i have ?

- anonymous

This is getting confusing for me :(

- anonymous

me too! haha. and yay '94 kids !

- anonymous

i am not sure what direction the force on point a is acting. perpendicular to the rod or the cable?

- S

Does the answer 12.78 lb sound reasonable? I can't remember how do you do this, so I just did a bunch of stuff that seemed to be reasonable and I got that =\

- anonymous

it does sound reasonable! but the back of the book has the answers, and evidently T = 9.08 lb ):

- S

Hmmm.. Well at first i did what you did, sum of the forces in the y direction:
Ay + Cy - T - 15 = 0

- S

A is experiencing a force T, that's pulling on it at 10 degrees, so the y component of that force T that is acting on a would be sin(10) that would be equal to 0.1736T.
So now 0.1736T + Cy - 1T - 15 = 0
-0.8264T + Cy - 15 = 0

- S

Now if we can substitute something like that for Cy as well we could move on, but idk what to substitute for it cause there are no degrees for point C

- anonymous

maybe we are overlooking something. ummmmm.

- anonymous

http://www.wolframalpha.com/input/?i=cos+40*15%2F%282*cos10*sin40%29

- anonymous

explain pleaseee.

- anonymous

\[T \times \cos 10 \times 5 \times \sin 40 = \frac{5}{2} \times \cos 40 \times 15\]
T is the tension in the cable.
I used concept of Torques, one Torque is in effect due to its weight (rod's weight) the other is from the horizontal component of Cable's Tension (T * cos 10).
Note. I calculated Toque about A.

- S

Sum of x direction
Ax - Fb = 0 Fb you said = 8.9
Ax - 8.9 = 0
Ax = 8.9
cos(10)T = 8.9
0.98T = 8.9
T = 9.08

- anonymous

i havent learned torques, but s is something it looks like i can understand. thank you both !!

- S

That kinda looks reasonable when you look at it. I think theres no Cx because its just like a continuation of Ax or something.. idk.. i can't remember.. I took it like a year and a half ago so i can't remember.. i'm gonna be a crappy engineer lol.. can't remember much lol

- S

But at least it give the right answer.. i hope the procedure is right as well =)

- anonymous

sike i still dont know what engineering even is and i am almost done with my first year of college. gracias yall !

- S

Don't worry, you'll figure it out =D
De nada =D

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